\subsection{Integration by parts}
We can use the fundamental theorem of calculus to deduce familiar integration techniques, such as integration by parts, and integration by substitution.
\begin{corollary}
	Suppose \(f', g'\) exist and are continuous on \([a, b]\).
	Then
	\[
		\int_a^b f'g = \eval{fg}_a^b - \int_a^b fg'
	\]
\end{corollary}
\begin{proof}
	By the product rule, we have
	\[
		(fg)' = f'g + fg'
	\]
	Then by the fundamental theorem of calculus,
	\[
		\int_a^b (fg)' = \eval{fg}_a^b = \int_a^b f'g + \int_a^b fg'
	\]
	and the result follows.
\end{proof}

\subsection{Integration by substitution}
\begin{corollary}
	Let \(g \colon [\alpha, \beta] \to [a, b]\) with \(g(\alpha) = a, g(\beta) = b\) and let \(g'\) exist and be continuous on \([\alpha, \beta]\).
	Let \(f \colon [a, b] \to \mathbb R\) be continuous.
	Then
	\[
		\int_a^b f(x)\dd{x} = \int_\alpha^\beta f(g(t))g'(t)\dd{t}
	\]
\end{corollary}
\begin{proof}
	Let \(F(x) = \int_a^x f(t) \dd{t}\).
	Then let \(h(t) = F(g(t))\).
	This is well defined since \(g\) takes values in \([a, b]\).
	Then,
	\begin{align*}
		\int_\alpha^\beta f(g(t))g'(t)\dd{t} & = \int_\alpha^\beta F'(g(t))g'(t) \dd{t} \\
		                                     & = \int_\alpha^\beta h'(t) \dd{t}         \\
		                                     & = h(\beta) - h(\alpha)                   \\
		                                     & = F(b) - F(a)                            \\
		                                     & = F(b)                                   \\
		                                     & = \int_a^b f(x) \dd{x}
	\end{align*}
\end{proof}
